Rolle’s theorem, mean value theorem and consequences

Goals and prerequisites

Use existence theorems only after checking their hypotheses. Prerequisites: continuity, derivatives and interval notation. All intervals below have real endpoints a<b.

1. Secant slope and tangent slope

The secant slope between (a,f(a)) and (b,f(b)) is [f(b)−f(a)]/(b−a). The mean value theorem (MVT) states: if f is continuous on [a,b] and differentiable on (a,b), at least one c∈(a,b) satisfies f′(c)=[f(b)−f(a)]/(b−a). Endpoints need continuity but do not need two-sided derivatives. The point c must be strictly inside the interval.

The theorem asserts existence, not uniqueness, and does not claim that c is the midpoint. To solve, first check the hypotheses, calculate the secant slope, solve the derivative equation, and retain only solutions in (a,b).

Example: f(x)=x² on [1,3]. The secant slope is (9−1)/2=4. Since f′(c)=2c, c=2 works. Example: f(x)=x³ on [0,2]. The slope is 4, so 3c²=4 and the admissible point is c=2/√3, not the midpoint 1 and not the negative root.

2. Rolle's theorem is the zero-slope case

If the MVT hypotheses hold and f(a)=f(b), at least one c∈(a,b) has f′(c)=0. For f(x)=x²−4 on [−2,2], both endpoint values are zero and c=0 satisfies the conclusion.

Why it works: continuity on the compact interval gives extrema. If f is constant, every interior derivative is zero. Otherwise, equal endpoint values force an absolute extremum at an interior point; differentiability and Fermat's condition give a zero derivative there. For general MVT, subtract the secant line: g(x)=f(x)−f(a)−[(f(b)−f(a))/(b−a)](x−a). Then g(a)=g(b)=0, so Rolle applied to g gives the required equation.

3. Hypotheses are not decoration

For |x| on [−1,1], endpoints agree and the function is continuous, but it is not differentiable at zero. Its derivative is −1 or 1 elsewhere, so no c has derivative zero. Rolle cannot be applied.

For f(x)=1/x on [−1,1], the function is not even defined at zero, so neither continuity nor differentiability holds throughout the required sets. One cannot use the endpoint values alone.

Failure of a hypothesis means that the theorem cannot guarantee the conclusion. It does not mean that the conclusion is necessarily false. A separate direct calculation may still find a suitable c.

4. From local derivatives to global bounds

If f is continuous on [a,b], differentiable on (a,b), and m≤f′(x)≤M for every interior x, MVT gives m(b−a)≤f(b)−f(a)≤M(b−a). If |f′|≤K, then |f(b)−f(a)|≤K|b−a|. For sin x, |cos x|≤1 gives |sin b−sin a|≤|b−a|.

If a function is differentiable throughout an interval and f′=0 everywhere, then f is constant on that interval: apply MVT between any two points. The interval requirement matters. A function can be separately constant with different values on (−∞,0) and (0,∞), yet have zero derivative everywhere in that disconnected domain.

Likewise, f′>0 throughout an interval implies strict increase. For root uniqueness, combine this with a separate existence argument. For f(x)=x³+x−1 on [0,1], continuity and endpoint signs give a root, while f′=3x²+1>0 gives strict increase and hence at most one root. Therefore exactly one root lies in (0,1).

5. Exam checklist

Write the closed interval for continuity and the open interval for differentiability. Verify equal endpoints only for Rolle. Reject boundary and out-of-domain derivative solutions. Never infer a unique c from MVT alone. Distinguish “the theorem applies” from “its conclusion happens to hold.”

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CM01. For f(x)=x² on [1,3], which c satisfies MVT?

A. 1 B. 2 C. 3 D. 4

Answer: B

A: An endpoint is not admissible and its slope is 2.

B: The secant slope 4 equals 2c at c=2.

C: An endpoint is not admissible and its slope is 6.

D: This is the secant slope, not c.

CM02. Why does Rolle not apply to |x| on [−1,1]?

A. Unequal endpoint values / सिरों के मान अलग B. Not continuous / सतत नहीं C. Not differentiable at 0 / 0 पर अवकलनीय नहीं D. Interval is closed / अंतराल बंद है

Answer: C

A: Both endpoint values are 1.

B: The function is continuous everywhere.

C: The left and right derivatives at zero disagree.

D: A closed interval is required for continuity.

CM03. Under MVT hypotheses on [a,b], c is guaranteed to lie in:

A. (a,b) B. {a,b} C. Only the midpoint / केवल मध्यबिंदु D. Outside [a,b] / [a,b] के बाहर

Answer: A

A: The theorem guarantees an interior point.

B: The endpoints are excluded.

C: The midpoint is not generally the required point.

D: The guaranteed point lies inside.

CM04. A differentiable f on an interval has f′=0 everywhere. Then f is:

A. Necessarily zero / अनिवार्यतः शून्य B. Strictly increasing / कठोर बढ़ता C. Strictly decreasing / कठोर घटता D. Constant / अचर

Answer: D

A: Any constant, not just zero, has zero derivative.

B: MVT gives equal values at any two points.

C: MVT gives equal values at any two points.

D: Between any two points MVT makes the change zero.

CM05. If f is continuous on [2,5], differentiable inside and 1≤f′≤4, which bound follows?

A. 1≤f(5)−f(2)≤4 B. 3≤f(5)−f(2)≤12 C. 2≤f(5)−f(2)≤5 D. f(5)−f(2)=3

Answer: B

A: Multiply the derivative bounds by interval length 3.

B: MVT gives f(5)−f(2)=3f′(c).

C: Endpoint coordinates are not derivative bounds.

D: Only a range is guaranteed, not equality.

CM06. How many real roots does x³+x−1 have in (0,1)?

A. 0 B. At least two / कम से कम दो C. Exactly one / ठीक एक D. Infinitely many / अनंत

Answer: C

A: Opposite endpoint signs and continuity give a root.

B: Positive derivative prevents two roots.

C: Continuity gives existence; 3x²+1>0 gives uniqueness.

D: Strict increase rules this out.

Source and scope

Official Mean value theorem heading; Rolle and derivative consequences support its use.

Official GATE 2026 CS syllabus (verified 2026-10-14).